Find all roots of the equation x4-2x^3+14x^2-18x+45=0

Quadratic factors of a quartic can be found with some trial and error. You can make use of the expansion ...

to write four equations in a, b, c, and d. These can be somewhat messy to solve, but some "guess and check" will get to a solution without too much trouble.

Here, the equations you get are ...

Since 45 has only 6 divisors, the last of these gives rise to 6 possibilities to check: (b, d) = (±1, ±45), (±3, ±15), and (±5, ±9).

We choose to check (b, d) = (5, 9) first. Then our equations reduce to

and it becomes pretty clear that c=0, a=-2 is a solution to these. Then our factorization is

The first quadratic can be written as

... (x-1)² +4 = 0

so has roots that can be found by subtracting 4, taking the square root, then adding 1.

... x = 1 ± √-4 = 1 ±2i

The second factor's roots can be found by subtracting 9 and taking the square root.

... x = √(-9) = ±3i

So the roots of the original quartic equation are ...

... {1 -2i, 1 +2i, -3i, +3i}

answer: the answer is b. i think hope i got it correct

step-by-step explanation: